Chapter 10: Gases
1. Gas Properties
2. Empirical Gas Laws
3. Ideal Gas Law
4. Gas Stoichiometry
5. Dalton’s Law of Partial Pressures
6. Kinetic Molecular Theory of Gases
7. Effusion & Diffusion
8. Real Gases
The Gaseous State
What are the Characteristics of the
Gaseous State?
1. Fluid - Will Take the Shape of a Container
2. Compressible - Density Changes As the
Volume Changes.
3. Forms Solutions (Homogeneous Mixtures)
How Can We Measure the Quantity
of Molecules in the Gaseous State?
Pressure Measurement
11.1 Pressure
Pressure is defined as the force per unit area.
The Barometer
F
P
A
Units of Pressure:
1atm = 760 torr
1atm = 760mm Hg
1atm = 14.7lb/in2
1atm = 1.01x105 Pa
1Pa = 1N/m2
= 1kg/(m.s2)
Evangelista Torricelli
(1608-1647)
Why Does the
Weatherman Report
the Barometric
Pressure?
The Manometer
Pg + Pc = Patm
Pg
PC
11.2 Empirical Gas Laws
Pg = Patm + Pc
Patm
Pg
Boyle’s Law, PV=k
Patm
Charle’s Law, V=bT
PC
Avogadro’s Law, V=an
@ const. n,T
@ const. n,P
@ const. P,T
1
Boyle’s Law Data
From Boyle’s Law, PV=k, What
Would You Graph?
V= K(1/P)
or
P=k(1/V)
Knowing V at one P, how do we
determine it at another P?
This is a “Two State” Problem
P1V1=k=P2V2
P1V1=P2V2
eq. of straight line
with y intercept = 0
Y=mx+b
V=k(1/P) + 0
Class Problem, If a sealed container at
atmospheric pressure is doubled in
size, what is the new pressure?
P2= P1(V1/V2)
Charles’s Law & Absolute Zero
Avogadro’s Hypothesis
What happens at pt. A?
0K = Absolute
Zero.
It has never
been obtained
A
Equal Volumes of Gases at the Same
Temperature and Pressure Have the
Same Number of Molecules
Negative Kelvin
temperature
implies negative
volume
Avogadro’s Law
-Effect of changing the number of
gaseous molecules on the volume
V=an @ const. P,T
What are the units of Avogadro’s Constant.
If 3 moles H2(g) occupies 2L, what
volume does 4 moles occupy?
This is a “Two State” Problem
V1=an1 and V2=an2
Dividing state one by the equality of state two
cancels Avagardo’s constant:
V1=an1
V2=an2
V1=n1
V2=n2
V1= V2
n1
n2
Solve algebraically first, then assign state
(State 1 is volume of 4 moles)
2
11.3 Ideal Gas Law: Eq of State
Summary of Gas Laws
PV=nRT
R=0.08206(l.atm)/(mol.K)
Note, from chapter 6,
(DH=E+PDV) and the
ideal gas law relates the
absolute temperature to
the molar work of
expansion. So R has
units of energy/mol-K.
Ideal Gas Law: PV = nRT
Boyle’s Law:
Charles’s Law:
k = nRT
V = bT,
b = nR/P
Avogadro’s Law: V = an,
a = RT/P
Ideal Gas Law Calculations
Group Problem: What volume
would 30.0g of H2O(g) occupy
at 280. oF and 400. torr?
PV = k,
STP
Standard Temperature & Pressure
Properties of gases are often tabulated at STP:
Standard Temperature = 0o C
Standard Pressure = 1 Atm
(actually 1 bar = .987 atm)
What is the Molar Volume of an Ideal
Gas at STP?
(Keep units in calculations)
Ideal Gas Molar Volume
nRT
V
P
V
(1mol )(.0826 L  atm mol  K )(273K )
1atm
Two State Approach to Ideal Gas
Calculations
P1V1=n1RT1
P2V2=n2RT2
P1V1=n1RT1
P2V2=n2RT2
=
P1V1=n1T1
P2V2=n2T2
V=22.4 L at STP
Interactive Quiz 12.4
3
Gas Density
Two State Eq. vs. Ideal Gas Laws
When would you want to use the Two State
Equation instead of the Ideal Gas Law?
d=m,
v
Use units of g/L (not g/mL)
How can we determine the density of a gas?
-Use Two State Eq. if given two sets of
conditions for the same gas
(say STP plus another)
The Problem is gasses are compressible
and so the density is not a constant value
like it is for solids and liquids.
-Use ideal gas law if given 3 of the 4
variables (P,V,n & T) or if n changes
-Need an equation of state like the Ideal
Gas Equation
Gas Density
d=m,
v
Use units of g/L (not g/mL)
The mass of a gas can be determined
from the # of moles (n) and molar mass
(formula weight, fw) of the gas.
Note,
Ideal Gas Density
PV = nRT
n(moles) = m(g)
fw(g/mol)
PV = mRT
fw
d
Molar Mass Problem
An unknown gas has a density of 4.08g/l
at 88oC and 1.4 atm, what is its molar
mass?
n = # moles
m = mass
fw = molar mass
m P( fw)

V
RT
Density Problem
What is the density of sulfur dioxide gas
at 88oC and 1.4 atm?
4
11.4 Gas Stoichiometry
Gas Stoichiometry
Consider the Decomposition of Potassium Chlorate
Knowing the P,V & T, gives the moles of a
gas, therefor we can use gas phase data
in stoichiometric calculations
2KClO3(s)
MnO2,D
2KCl(s) + 3O2(g)
How many grams KClO3 would produce enough
O2(g) to occupy 30mL at 1 atm and 68oF.
1. Gas Phase Step - Determine moles O2
needed
2. Stoichiometric Step - Determine mass
of KClO3 required to produce the O2.
1. Gas Phase Calculations
n = PV
RT
T ( K )  273.15 C  [
2. Stoichometric Calculations
2KClO3(s)
MnO2,D
2KCl(s) + 3O2(g)
(68 F  40)
 40] C  293.15K
1.8
How Many Liters of Hydrogen Gas (at
35o C & 720 torr) Will Be Released
When 200.0 mL of 3.00 M HCl reacts
with 10.0 g Mg?
2. Determine Moles of Each Reactant
1. This is a Single Replacement Reaction
3. Determine Limiting Reagent
Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g)
-Note how this reaction equation involves
species in all three phases
5
Calculate Volume of Hydrogen Gas
12.5 Dalton’s Law of Partial Pressure
For a mixture of gases, the total
pressure is the sum of the
pressures due to each type of gas
Partial Pressure - is the pressure due to
a specific type of gas
Total Pressure
P
T
PT  n1 (
RT
RT
RT
)  n2 (
)  n3 (
)
V
V
V
PT  (n1  n2  n3 )(
PT  nT
RT
V
RT
)
V
Calculate
i
i th gas
XA is the mole fraction of species A,
where XA = nA
nT
PA
PT
=
nA(RT/V)
nT(RT/V)
=
nA
= XA
nT
PA = XAPT
Were nT = n1 + n2 + n3
A Tank Contains 78.0 g of N2 and 42 g of
He at a total pressure of 3.75 atm and a
temperature of 50.0oC.
P
Partial Pressure
of ith gas
Mole Fraction
Consider a mixture of 3 gases, 1, 2 & 3
PT = P1 + P2 +P3

all gasses
T(oC) P(torr)
10
20
25
30
40
60
70
90
100
9.209
17.535
23.756
31.824
55.324
149.4
233.7
525.8
760
Gasses Over Water
1. What Is the Partial Pressure
of Oxygen If the Gas at 25oc
Has a Pressure of 720. Torr?
1. Moles N2
2. Moles He
PT= Poxygen+Pwater=
3. Partial Pressure of N2
Poxygen= 720-23.756
4. Partial Pressure of He
=696torr
6
T(oC) P(torr)
10
20
25
30
40
60
70
90
100
9.209
17.535
23.756
31.824
55.324
149.4
233.7
525.8
760
Gasses Over Water
2. If 0.50 L of Water Is
Displaced, What Mass of
Oxygen Was Produced?
12.6 Kinetic Molecular Theory of
Gases (KMT)
1. The Volume of Gas Particles are Negligible Compared to
the Total Gas Volume; Most of the Volume is Empty
Space
2. Gas Particles are in Constant Motion, Pressure Results
from Elastic Collisions with the Walls of the
Container (the particles do not lose energy)
3. The Average Kinetic Energy of Gas Particles are
Proportional to the Kelvin (Absolute) Temperature
4. There Are Negligible Intermolecular Forces
KMT and P,T
P = (Constant)T b = nR
V
As the temperature increases
the kinetic energy increases
and thus the force of impact
with the container increases
causing P to increase
T(oC) P(torr)
10
20
25
30
40
60
70
90
100
9.209
17.535
23.756
31.824
55.324
149.4
233.7
525.8
760
Gasses Over Water
1. What mass of KClO3 was
consumed when the oxygen
produced displaced 20.0L of
water at 25.0oC & 720. torr?
KMT and Boyle’s Law (P,V)
P = (K)1
V
K = nRT
As the volume gets smaller
the mean free path decreases
and the collision frequency
goes up, therefore P increases
as V decrease
KMT and Charle’s Law (V,T)
V = (b)T
b = nR
P
As the temperature increases
the pressure tends to increase
but since this is at constant
pressure, the volume
increases to compensate
7
KMT and Avogadro’s Law (V,n)
V = (a)T
a = RT
P
As the number of molecules
increases, the collision
frequency increases causing
the volume to increase at
constant pressure
Kinetic Energy
EK = The Energy of Motion
For a Particle (molecule)
EK = 1 (mass)(speed)2 = 1 mv2
2
2
What Variables Effect the
Kinetic Energy?
KMT & Temperature
Kinetic Energy = 1/2mv2
How will increasing
the temperature effect
the distribution of
molecular speeds?
KMT and Dalton’s Law of Partial
Pressures
Since there are no
intermolecular interactions
the total pressure is
proportional to the total
number of molecules
regardless of their identity
The Absolute Temperature (K)
Not Molecular Size!
Larger Molecules Have a
Lower Average Velocity
Note, a small gas particle like H2 has the
same average kinetic energy as a large
particle like Cl2, (H2 having the smaller
mass, has the greater average velocity)
KMT & Temperature
How does the molecular
size influence the
velocity distribution at
constant Temperature?
Heaviest Molecules
Low T
High T
Does the area under the
curve change as the
temperature changes?
Middle Weight Molecules
Lighter Molecules
8
Root Mean Square Velocity
Note velocity is a vector and in 3D
space the average velocity of random
motion is zero as the components in
opposite directions cancel.
m2  mx2 + my2 + mz2
For an ensemble of molecules
KE = Ave Kinetic Energy
m  root mean
Ek ,ave
m
mrms  m  m  m
2
x
2
y
2
z
KE = 1 mu2
square velocity
2
 N A ( 12 mm 2 )  23 RT
m
, noting mNa  M (molar mass)
3 RT
mN a
3 RT
M
Graham’s Law:
Diffusion: Process where gas mix through
random motion
Effusion: Process of collision less
movement of gasses into a vacuum
Graham’s Law: Rate of Effusion is
Inversely Proportional to Molar Mass
Real Gasses:
When Do Real Gasses Deviate From Ideal?
-When the Postulates of the Kinetic
Molecular Theory Break Down
-High Pressure: Gas Molecules Take up
Substantial Space of Container and Being
Closer Together, Have Greater
Intermolecular Interactions
-Low Temperature: The Average Kinetic
Energy Is Decreased Causing a Greater
Influence of Intermolecular Forces on the
Gas Species Trajectory
Graham’s Law:
Ek  32 RT  N A 12 mm 2
Note:
At Constant Temperature both gases have same
kinetic energy

1
2
mm 2

gas1


1
2
mm 2
m m  m2 m
2
1 1
m1 m 22  m 2 

 
m2 m12  m1 
2
and

gas2
2
2
m2
m1
m2


m2
m1
m1
Non Ideal Gases
Van Der Waals Equation
n
[ P  a( )2 ][V  nb]  nRT
V
Correction Term for
Intermolecular Forces
Correction Term for
Molecular Volume
9
Van Der Waals Equation
n
[ P  a( )2 ][V  nb]  nRT
V
Gas
a (atm-L2)
mol2
b (L)
mol
He
Xe
CH4
0.0341
4.19
2.25
0.0237
0.0511
0.0428
NH3
4.17
0.0371
H2O
5.46
0.0305
10